\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

\(B=\left(4x^2+4x+1\right)-3\)

\(B=\left(2x+1\right)^2-3\)

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

\(\Rightarrow\left(2x+1\right)^2-3\ge-3\)

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(P=x^2-4x+4+y^2-6y+9-8\)

\(=\left(x-2\right)^2+\left(y-3\right)^2-8\ge-8\)

vậy GTNN của P là -8 khi \(x=2;y=3\)

A=\(\left(x-y\right)^2-2.6.\left(x-y\right)+36+5y^2+10y+5+4\)

=\(\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

Dấu bằng xảy ra khi y=1 và x=5

2B=\(2x^2+2y^2-2xy-2x+2y+2\)

=\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

=>B\(\ge\)0

Câu b mình viết nhầm dấu \(\ge\)đáng lẽ đúng phải là \(\le\)

a)

\(A=x^2+y^2-x+6y+10.\)

\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy \(MinA=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}}\)

b)

\(B=2x-2x^2-5\)

\(=-2\left(x^2-x+\frac{1}{4}\right)+2.\frac{1}{4}-5\)

\(=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Vậy \(MaxB=-\frac{9}{2}\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

C = x² + 4x + y² - 6y + 11 ( sửa -y² => +y² chứ để như kia không tìm được :)) )

= ( x² + 4x + 4 ) + ( y² - 6y + 9 ) - 2

= ( x + 2 )² + ( y - 3 )² - 2 ≥ -2 ∀ x, y

Đẳng thức xảy ra <=> x = -2 ; y = 3

=> MinC = -2 <=> x = -2 ; y = 3

Sửa đề C = - x² - 4x - y² - 6y + 11

<=> C = - ( x² + 4x + 4 ) - ( y² + 6y + 9 ) + 24

<=> C = \(-\left(x+2\right)^2-\left(y+3\right)^2+16\le16\)

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}-\left(x+2\right)^2=0\\-\left(y+3\right)^2=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\y=-3\end{cases}}\)

Vậy maxC = 24 <=> x = - 2 ; y = - 3

1. a. \(A=8a-8a^2+3=-8\left(a-\frac{1}{2}\right)^2+5\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow-8\left(a-\frac{1}{2}\right)^2+5\le5\)

Dấu "=" xảy ra \(\Leftrightarrow-8\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)

Vậy Amax = 5 <=> a = 1/2

b. \(B=b-\frac{9b^2}{25}=-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\)

Vì \(\left(b-\frac{25}{18}\right)^2\ge0\forall b\)\(\Rightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\le\frac{25}{36}\)

Dấu "=" xảy ra \(\Leftrightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2=0\Leftrightarrow b-\frac{25}{18}=0\Leftrightarrow b=\frac{25}{18}\)

Vậy Bmax = 25/36 <=> b = 25/18

a,\(A=8a-8a^2+3\)

       \(=-8\left(a^2-a\right)+3\)

       \(=-8\left(a^2-2a\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)+3\)

       \(=-8\left[\left(a-\frac{1}{2}\right)^2-\frac{1}{4}\right]+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+2+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+5\le5\forall a\) 

Dấu"=" xảy ra khi \(\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)

Vậy \(Max_A=5\)khi\(a=\frac{1}{2}\)

bài 2:

b,\(D=d^2+10e^2-6de-10e+26\)

\(=d^2-23de+\left(3e\right)^2+e^2-2.5e+5^2+1\)

\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)

Dấu"=" xảy ra khi\(\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}}\)

vậy \(D_{min}=1\)khi \(d=15;e=5\)

c,:\(E=4x^4+12x^2+11\)

\(=\left(2x^2\right)^2+2.2x^2.3+3^2+2\)

\(=\left(2x^2+3\right)^2+2\ge2\forall x\)

còn 1 đoạn nx bạn tự lm tiếp,lm giống như D

a)\(A=4x^2+4x+11\)

\(=4x^2+4x+1+10\)

\(=\left(2x+1\right)^2+10\ge10\)

Dấu = khi \(x=\frac{-1}{2}\)

Vậy MinA=10 khi \(x=\frac{-1}{2}\)

b)\(B=3x^2-6x+1\)

\(=3x^2-6x+3-2\)

\(=3\left(x^2-2x+1\right)-2\)

\(=3\left(x-1\right)^2-2\ge-2\)

Dấu = khi \(x=1\)

Vậy MinB=-2 khi \(x=1\)

c)\(C=x^2-2x+y^2-4y+6\)

\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

1. D = 3( x² - 2x.1/3 + 1/9) -1/3 +1

GTNN D = 5/6

dài quá, nản quá

tks bn

Bài 1

\(A=7^6.2^6-\left(14^3+5\right)\left(14^3-5\right)\\ A=\left(7.2\right)^6-\left(14^6-25\right)\\ A=14^6-14^6+25\\ A=25\)

Vậy A = 25

Cảm ơn nha bn bít giải mấy bài khác ko,giúp mk vs